About $\int_{-\infty}^{0}\text{arcsin}\left(e^x\right)\text{d}x$ I've shown that the integral $I$ exists
$$I=\int_{-\infty}^{0}\text{arcsin}\left(e^x\right)\text{d}x \approx 1.089$$
Is there a way to find the exact value ?
Using $u=e^x$ we have $\text{d}u=u\text{d}x$ so
$$
I=\int_{0}^{1}\frac{\text{arcsin}\left(u\right)}{u}\text{d}u
$$
But it did not help me.
 A: Applying IBP on your last integral followed by the substitution $u\mapsto \sin(u)$ we obtain that
\begin{align*}
I=\int_0^1\frac{\arcsin(u)}u\mathrm du&=\underbrace{\left[\log(u)\arcsin(u)\right]_0^1}_{\to 0}-\int_0^1\frac{\log(u)}{\sqrt{1-u^2}}\mathrm du\\
&=-\int_0^{\pi/2}\log(\sin u)\mathrm du
\end{align*}
There are various way to evaluate the latter integral. I will use for the sake of simplicity the Fourier Series Expansion of $\log(\sin u)$ here. Thus, we further get that
\begin{align*}
I=-\int_0^{\pi/2}\log(\sin u)\mathrm du&=-\int_0^{\pi/2}\left[-\log(2)-\sum_{k=1}^\infty \frac{\cos(2ku)}{k}\right]\mathrm du\\
&=\frac\pi2\log(2)+\int_0^{\pi/2}\sum_{k=1}^\infty \frac{\cos(2ku)}k\mathrm du\\
&=\frac\pi2\log(2)+\sum_{k=1}^\infty\frac1k\int_0^{\pi/2}\cos(2ku)\mathrm du\\
&=\frac\pi2\log(2)+\sum_{k=1}^\infty\frac1k\underbrace{\left[\frac{\sin(2ku)}{2k}\right]_0^{\pi/2}}_{=0}
\end{align*}

$$\therefore~I~=~\int_0^1\frac{\arcsin(u)}u\mathrm du~=~\frac\pi2\log(2)$$

Note that one could also invoke the Clausen Function aswell as the first derivative of the Beta Function in order to evaluate the logarithmo-trigonometric integral. 
A: Next write $y=\arcsin u$ so $\mathrm{d}u=\cos y\mathrm{d}y$ and $I=\int_0^{\pi/2}y\cot y\mathrm{d}y$. This is a famous integral. One way to compute it is with Feynman's trick, viz. $$I(k)=\int_0^{\pi/2}\frac{\arctan (k\tan y)}{\tan y}\mathrm{d}y=\int_0^k\mathrm{d}k^\prime\int_0^{\pi/2}\frac{\mathrm{d}y}{1+k^{\prime2}\tan^2 y}\\=\int_0^k\mathrm{d}k^\prime\int_0^{\infty}\frac{{d}t}{(1+t^2)(1+k^{\prime2}t^2)}.$$The rest is an exercise in partial fractions, eventually obtaining$$I(k)=\frac{\pi}{2}\int_0^k\frac{\mathrm{d}k^\prime}{1+k^\prime}=\frac{\pi}{2}\ln|1+k|.$$The case $k=1$ agrees with mrtaurho's answer.
A: Use $\;u:=\sin(t)\;$ to continue and integrate by parts to get :
\begin{align}
I&=\int_0^{\large{\frac {\pi}2}}\frac t{\sin(t)}\cos(t)\,dt\\
&=\left.t\,\log(\sin(t))\right|_0^{\large{\frac {\pi}2}}-\int_0^{\large{\frac {\pi}2}}\log(\sin(t))\,dt\\
&=\frac{\pi}2\log(2)\\
\end{align}
since the last integral is well known at SE (for example here) 
